# If $S \times \Bbb{R}^k$ is homeomorphic to $T \times \Bbb{R}^k$ and $S$ is compact, can we conclude that $T$ is compact?

If $S \times \Bbb{R}^k$ is homeomorphic to $T \times \Bbb{R}^k$ for some $k \geq 1$ and $S$ is compact, can we conclude that $T$ is compact?

The case where $k=1$ and $S$ and hence also $T$ are locally connected is dealt with in my answer to this question. I am now curious about the general case.

• The choice of $S$ and $T$ is unfortunate. I first read them as $S$ for the circle and $T$ for the torus. Are they just random spaces? – Fly by Night May 8 '13 at 15:40
• Yes, they are arbitrary. – Alexander Thumm May 8 '13 at 15:51
• It feels to me like some kind of cohomology at infinity calculation could show this, but I don't actually see an argument. – Cheerful Parsnip May 8 '13 at 17:41
• I can't even think of two non-homeomorphic bases such that the cylinders are homeomorphic... The only thing that is obvious to me is that connectedness of the basis is invariant. Nice question. – Julien May 8 '13 at 17:45
• @julien: Bing's dogbone space and the Whitehead manifold are both spaces which become homeomorphic to $\mathbb R^4$ when you cross with $\mathbb R$. The first is not a manifold and the second one is a $3$-manifold. Such things happen a lot in higher dimensions. – Cheerful Parsnip May 8 '13 at 18:07

Yes, if $S$ is compact, then $T$ is also compact.

In fact, the property that $S$ is compact can be expressed in terms of the topological space $X=S\times\mathbb{R}^ k$. The idea is natural enough, if $S$ is compact then it is possible to remove a compact set from $X$ which stops certain maps from the $(k-1)$-sphere $\mathbb{S}^{k-1}$ from being null-homotopic. The precise statement is a little tricky though.

The following are equivalent.

1. $S$ is compact.
2. There exists a compact $A\subseteq X$ such that, for every compact $B\supseteq A$ there exists a compact $C\supseteq B$ such that, for every $x\in X\setminus C$, there exists a continuous $f\colon \mathbb{S}^{k-1}\to X\setminus B$ containing $x$ in its image and which is null-homotopic as a map to $X$ but not null-homotopic as a map to $X\setminus B$.

First, suppose that 1 is true. We need to construct the sets $A$, $C$ and the function $f$ for any choices of $B$ and $x$. Set $A=S\times\lbrace0\rbrace$. Given any compact $B\supseteq A$, let $\pi_{\mathbb{R}^k}(B)$ be its projection onto $\mathbb{R}^k$. As this is compact, it is contained in the closed ball $\bar B_R(0)$ for some $R > 0$. Set $C=S\times\bar B_R(0)$. Then, given any $x\in X\setminus C$, we can write $x=(s_0,y_0)$ where $\lVert y_0\rVert = r > R$. Define $f\colon \mathbb{S}^{k-1}\to X$ by $f(a)=(s_0,ra)$. This is null-homotopic as a map to $X$ via the homotopy $F(a,u)=(s_0,(1-u)ra)$. On the other hand, it is not null-homotopic as a map to $X\setminus B$. To see this, define $g\colon X\setminus B\to \mathbb{S}^{k-1}$ by $g(s,y)=y/\lVert y\rVert$. Then, $g\circ f$ is the identity on $\mathbb{S}^{k-1}$. However, it is standard that the (k-1)-sphere is not contractible or, equivalently, the identity on $\mathbb{S}^{k-1}$ is not null-homotopic, implying that $f$ is not null-homotopic.

Conversely, suppose that $S$ is not compact. We need to prove that 2 is false. To be explicit, the negation of 2 can be written as

2'. For any compact $A\subseteq X$ there exists a compact $B\supseteq A$ such that, for any compact $C\supseteq B$, there exists an $x\in X\setminus C$ such that every continuous $f\colon\mathbb{S}^{k-1}\to X\setminus B$ with $x$ in its image and which is null-homotopic as a map to $X$ is also null-homotopic as a map to $X\setminus B$.

So, we need to construct the set $B$ and point $x\in X\setminus C$ for any choice of $A$ and $C$, and show that any such function $f$ is null-homotopic as a map to $X\setminus B$.

Given a compact $A\subseteq X$, its projection onto $\mathbb{R}^k$ is compact, so contained in the closed ball $\bar B_R(0)$ for some $R > 0$. Also, its projection $S_1=\pi_S(A)$ is compact. Take $B=S_1\times\bar B_R(0)$. If $C$ is a compact set containing $B$ then the projection $\pi_S(C)$ is compact, so is not the whole of $S$. Therefore, we can choose $s_0\in S\setminus\pi_S(C)$. Set $x=(s_0,0)\in X\setminus C$.

Now, suppose that $f\colon\mathbb{S}^{k-1}\to X\setminus B$ has $x$ in its image and is null-homotopic as a map to $X$. We need to construct a null-homotopy for $f$ as a map to $X\setminus B$. Write $f(a)=(f_1(a),f_2(a))$ where $f_1,f_2$ are maps from $\mathbb{S}^{k-1}$ to $S$ and $\mathbb{R}^k$ respectively. First, there will exists continuous maps $f^\prime_2\colon\mathbb{S}^{k-1}\to\mathbb{R}^k\setminus\lbrace0\rbrace$ which are arbitrarily close to $f_2$. One way to see this is to first approximate $f_2$ by a smooth function. Then, as the image of a smooth map from $\mathbb{S}^{k-1}$ to $\mathbb{R}^k$ cannot contain any open ball about the origin (its image must have zero measure), there will be arbitrarily small $v\in\mathbb{R}^k$ such that $f^\prime_2=f_2+v$ does not pass through the origin. Then, $f$ will be homotopic to $(f_1,f_2^\prime)$ for $f_2^\prime$ close enough to $f_2$. Next, by scaling, $f$ is homotopic to $(f_1,rf^\prime_2)$ for any $r\ge1$. So, by scaling $f_2^\prime$, we can assume that $\lVert f^\prime_2\rVert > R$.

We have shown that $f$ is homotopic to $(f_1,f^\prime_2)$ where $f^\prime_2\colon\mathbb{S}^{k-1}\to\mathbb{R}^{k-1}\setminus\bar B_R(0)$. Next, by projection onto $S$, the condition that $f$ is null-homotopic as a map to $X$ implies that $f_1$ is null-homotopic. Also, as $f$ has $(s_0,0)$ in its image, $f_1$ has $s_0$ in its image so is homotopic to the constant map to $s_0$. This gives a homotopy of $f$ to the map $a\mapsto(s_0,f^\prime_2(a))$. Finally, by scaling, $f^\prime_2$ is homotopic to the constant map to $0$. So, we have constructed a null-homotopy for $f$ (as a map to $X\setminus B$) to the constant map $a\mapsto(s_0,0)$.